To determine how far the shortstop needs to throw the ball to first base, we can envision the baseball diamond as a square. The bases are positioned at the corners of this square:
- Second base (2B) lies at (0, 90) feet
- First base (1B) lies at (90, 90) feet
- The shortstop is standing between second and third base (3B), which is at (0, 60) feet allowing her to be 30 feet away from second base.
Given that the shortstop is 30 feet away from second base, we can place her coordinates based on the location of second and third bases. Therefore, if she is 30 feet towards third base, her coordinates are:
- Shortstop (SS): (0, 60)
Now, first base (1B) has coordinates:
- First base (1B): (90, 90)
To find the distance from the shortstop to first base, we can use the distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\):
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Plugging in the coordinates of the shortstop and first base:
- \(x_1 = 0, y_1 = 60\) (shortstop)
- \(x_2 = 90, y_2 = 90\) (first base)
We have:
\[ \text{Distance} = \sqrt{(90 - 0)^2 + (90 - 60)^2} \]
Calculating this step-by-step:
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Subtract the coordinates:
- \(90 - 0 = 90\)
- \(90 - 60 = 30\)
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Square the differences:
- \(90^2 = 8100\)
- \(30^2 = 900\)
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Sum the squares:
- \(8100 + 900 = 9000\)
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Take the square root:
- \(\sqrt{9000} \approx 94.868\)
Rounding this to the nearest tenth gives approximately:
\(\text{Distance} \approx 94.9\) feet.
Thus, the shortstop needs to throw the ball approximately 94.9 feet to first base.
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